What is the integral from 1 to infinity of 9e^{-9x} ?

The integral from 1 to infinity of 9e^{-9x} is 1/(e^9)

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integral from 1 to infinity of 9e^{-9x}

\int_{1}^{\infty} 9 e^{- 9 x} d x

\frac{1}{e ^{9}}

Decimal

0.00012 \dots

Solution steps

\int_{1}^{\infty} 9 e^{- 9 x} d x

Compute the indefinite integral:

\int 9 e^{- 9 x} d x = - e^{- 9 x} + C

Compute the boundaries:

\int_{1}^{\infty} 9 e^{- 9 x} d x = 0 - (- \frac{1}{e ^{9}})

= 0 - (- \frac{1}{e ^{9}})

Simplify

= \frac{1}{e ^{9}}

10 y^{^{'}} + 20 y = 12

2 (12) (18) + 2 (9) \frac{d y}{d t} = 0

\int cos (2 x) sin (4 x) d x

\frac{d y}{d t} = 25 t^{4}

\frac{\frac{d y}{d t} - e ^{- t} + 6}{y} = - 6, y (0) = - 2

What is the integral from 1 to infinity of 9e^{-9x} ?
The integral from 1 to infinity of 9e^{-9x} is 1/(e^9)